
#1
Oct2708, 03:59 AM

P: 134

HI, I encountered a problem about the curve ball (table tennis ball) flying in the air. See the pictures below.
A ball with an initial velocity V along X axis, and has a spin along Z axis. As far As I know ( see my previous post in The air drag problem... ) . F1 is the air drag. F2 is the gravity. F3 is the magnus force( sometimes we call it lift force). So, the ball will not travel directly through X axis, it will curve to left. My question is : What about the spin axis? Will it shift from A1 to A2? ( A1 is the axis parallel to Z axis as the initial state )? If Yes, How can I estimate the angle between A1 and A2? Any suggestion is appreciated. Thank you very much. 



#2
Oct2708, 10:10 AM

P: 134

I'm sorry that the attachment image is destroyed(I upload the jpg file as an attachment, but it doesn't appeared). I will upload it soon!
Image added! 



#3
Oct3008, 04:53 AM

P: 4,513

The way you have it drawn, the ball will curve to the right, and also lift from gravitational effects.




#4
Oct3008, 05:38 AM

P: 134

curve ball problem, the spin axis will change?
Thanks Phark.
To my knowledge, I think the revolution axis(or spin axis) will remain its original direction( in the picture, it is A1), because, the drag, life and gravity force all have no effects on the spin Torque. 



#5
Oct3108, 04:44 AM

P: 4,513

My mistake about the directions of deflection. I got everything inverted.
The question is, what happens when the spin axis is at some accute angle to the velocity, like when the ball starts dropping, right? The lift and the induced drag, for instance, will act somewhere toward the 'leading edge'. But where is the profile dragthe viscous and eddy current drag? 



#6
Oct3108, 09:48 AM

P: 134

Hi, Phrak, I received your message. In my case: the revolution axis is transverse with Velocity, which is more simple than your question" spin axis is some accute angle to the velocity".
I think the lift force ( magnus force ) is defined by: [tex]\overrightarrow{F_{lifg}}=k1\overrightarrow{\omega}\times\overrightarro w{V}[/tex] Through I'm not sure this formula is right. The profile drag(Let me guess its meaning) is defined by: [tex]\overrightarrow{F_{drag}}=k2\parallel\overrightarrow{V}\parallel\overrightarrow{V}[/tex] where K1 and k2 are all drag coefficients. 



#7
Oct3108, 05:18 PM

P: 4,513

Let me make sure I have this right. You want to know if the spin axis will precess, right?
For something with little angular mass, such as a ping pong ball, it seems, it precession could be substancial at the right angle of attack. What's required for it to precess is force couple whos axis isn't parallel with the spin axis. I'd have to get out my crayons to see how the distributed forces would do this. It's probably more intuitive to draw the spin axis as horizontal, where the axis is oriented in the airflow like the span of a wing, then give it a little yaw. 



#8
Oct3108, 08:35 PM

P: 134

Thanks, Yes, I'm try to find whether the spin axis will precess.
Do you mean "Angular mass" is the same as "Angular momentum" ? see in http://en.wikipedia.org/wiki/Angular_momentum I'm my picture, I think the torque given by "distributed forces" is symmetric on the spin axis. So, the precess may not happen. 



#9
Nov108, 03:28 AM

Sci Advisor
P: 2,507

First of all, let me say that your direction of curve in the original illustration is correct. The ball will curve in the same direction it is spinning. The curve in the opposite direction would be true on Mars (or maybe at extremely high altitudes), but not on Earth.
Secondly, I can see no force that would cause the axis to tilt. As I understand it, precession is a phenominon that takes place when the axis of rotation is already tilted. Maybe I'm not using the right frame of reference? 



#10
Nov208, 12:05 AM

P: 134

Thanks LURCH for your reply.




#11
Nov208, 01:56 AM

P: 4,513





#12
Nov208, 01:10 AM

P: 4,513

It's almost a given that there will be precessional forces on the ball for the spin axis no at 0 or 90 degrees to the velocity. This would be false if there is some symmetry condition or the distributed forces somehow conspired against it.
The forces on the ball in this general condtion can be resolved into a force acting through the center of the ball (gravity included) and a force couple whos axis is not likely aligned with the spin axis. Any misalignment results in a pair of precessional forces. After breaking a few crayons trying some picturorial solutions, I've concluded wind tunnel simulation software might be better. In another senario, in the absense of gravity, and the spin axis not parallel nor perpendicular with the initial velocity, I think the ball will follow a corkscrew path; no precession required. 



#13
Nov208, 07:28 AM

P: 134

HI, Phark.
Firstly, I think the gravity will not give any precession effort on the initial Angular momentum. Secondly, If the initial angular momentum is parallel or perpendicular to the initial velocity(consider the absence of gravity condition), after a while, the axis will not change it's direction. I don't think it will get a "corkscrew path" ( the angular momentum is not parallel nor perpendicular to the velocity), though I can't test it in wind tunnel equipment. 


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